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    Question

    Suppose that the reliability of a COVID test is

    specified as follows: Of people having COVID, 90% of the test detect the disease but 10% go undetected. Of people free of COVID, 99% of the test are judged COVID–ive but 1% are diagnosed as showing COVID+ive. From a large population of which only 0.1% have COVID, one person is selected at random, given the COVID test, and the pathologist reports him/her as COVID+ive. What is the probability that the person actually has COVID?
    A 0.183 Correct Answer Incorrect Answer
    B 0.093 Correct Answer Incorrect Answer
    C 0.083 Correct Answer Incorrect Answer
    D 0.053 Correct Answer Incorrect Answer

    Solution

    Let E denote the event that the person selected is actually having COVID and A the event that the person's COVID test is diagnosed as +ive. We need to find P(E|A). Also, E’ denotes the event that the person selected is actually not having COVID. Clearly, {E, E'} is a partition of the sample space of all people in the population. We are given that P(E) = 0.1% = 0.1/100 = 0.001 P(E') = 1 – P(E) = 0.999 P(A|E) = P(Person tested as COVID+ive given that he/she is actually having COVID) = 90% = 90/100 = 0.9 and P(A|E') = P(Person tested as COVID +ive given that he/she is actually not having COVID) = 1% = 1/100 = 0.01 Now, by Bayes' theorem P(E|A) = [P(E) × P(A|E)]/[P(E) × P(A|E) + P(E') × P(A|E')] = [0.001 × 0.9]/[0.001 × 0.9 + 0.999 × 0.01] = 90/1089 = 0.083 approx.

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